Relativistic centripetal force

[starthaus]

It means the same thing as

$$\frac {d^2x'}{dt'^2}$$

when dx' = dx, as it does in this case.

:lol:

 

[yuiop]

$$\frac {d^2x}{dt'^2}$$ means nothing.

which is unfortunate, because that is exactly what you derive in the final expression (6) in your blog.

 

[starthaus]

which is unfortunate, because that is exactly what you derive in the final expression (6) in your blog.

Nope. $$\frac{d^2x}{dt^2}$$. New glasses, perhaps?

 

[yuiop]

Nope. $$\frac{d^2x}{dt^2}$$. New glasses, perhaps?

You effectively derive:

$$\frac{d^2x}{dt^2} = \gamma^{-2}\frac{d^2x}{dt'^2}$$

in expression (6) of your attachment, although you probably don't realize that.

 

[starthaus]

You effectively derive:

$$\frac{d^2x}{dt^2} = \gamma^{-2}\frac{d^2x}{dt'^2}$$

I would never write such frame-mixing nonsese.

in expression (6) of your attachment, although you probably don't realize that.

You definitely need to learn how to read math :

$$\frac{d^2x}{dt^2} = \gamma^{-2}(\frac{d^2x'}{dt'^2}+...$$

Can you see the pairing $$x,t$$ in the LHS and the pairing $$x',t'$$ in the RHS?
If you were less obsessed with finding errors where there are none, maybe you would be more able to learn.

 

[yuiop]

You effectively derive:

$$\frac{d^2x}{dt^2} = \gamma^{-2}\frac{d^2x}{dt'^2}$$

I would never write such frame-mixing nonsese.

You wrote such frame-mixing nonsense at step (5) when you derive the velocity transformation as:

$$\frac{dx}{dt} = \gamma^{-1}\frac{dx}{dt'}$$

 

[starthaus]

You wrote such frame-mixing nonsense at step (5) when you derive the velocity transformation as:

$$\frac{dx}{dt} = \gamma^{-1}\frac{dx}{dt'}$$

You definitely need a new pair of glasses :


$$\frac{dx}{dt} = \gamma^{-1}(\frac{dx'}{dt'}+...$$

Are you getting that desperate to prove me wrong that you can't even follow simple arithmetic anymore?

 

[Jorrie]

You wrote such frame-mixing nonsense at step (5) when you derive the velocity transformation as:

$$\frac{dx}{dt} = \gamma^{-1}\frac{dx}{dt'}$$

I have a feeling that the reason behind this 'brutal' argument is a miscommunication. Starthaus's full equations treat the general solution of rotating to Cartesian coordinate transformations. In this case, saying that dx' and dx are equivalent is not valid.

Kev and myself were mainly interested in one point only, say when omega=0, y=0 and dx' and dx are momentarily equivalent. If we have the magnitude of the proper acceleration at that point, we have it for all time (e.g., the reading of the mass on the scale against the inside wall of the rotating cylinder will remain constant). That way we simplify the route to a simple result.

 

[starthaus]

I have a feeling that the reason behind this 'brutal' argument is a miscommunication. Starthaus's full equations treat the general solution of rotating to Cartesian coordinate transformations. In this case, saying that dx' and dx are equivalent is not valid.

Good, you are making the effort to understand rather than find fault, like kev.

Kev and myself were mainly interested in one point only, say when omega=0,

$$\omega$$ cannot be ever zero, the frame is totating.

dx' and dx are momentarily equivalent.

This never happens. Look at the expression of dx, it is a function of (dx',dy',t')

There is something that you can do , though. You can consider the point (0,0) in the frame S'. When you do that you will obtain the correct expression for the transformation between frames but you need to do the math correctly, without attempting ugly hacks like setting dx=dx'.

 

[Jorrie]

$$\omega$$ cannot be ever zero, the frame is totating.

Sorry, I actually meant when $\theta = 0$, where $\omega=d\theta/dt$

It is perhaps time that you clear up your attachment - the first line of your equations [5] of "Uniformly Rotating Frames" does use the contentious 'mixing of frames' (dx/dt').

It would also be immensely helpful to all around here (especially me as an engineer, no mathematician) if you would correct the typo and complete equations [6]...

 

[yuiop]

$$\omega$$ cannot be ever zero, the frame is totating.

Jorrie obviously meant that when the angle (theta or phi or whatever you choose to call it) between the x axes of the two frames is zero and by the definition given in this link that you gave http://arxiv.org/PS_cache/gr-qc/pdf/9904/9904078v2.pdf this occurs when t' =0.

dx' and dx are momentarily equivalent.

This never happens. Look at the expression of dx, it is a function of (dx',dy',t')

It does happen. At time t' = 0, the axes are aligned and x = x' = 0 and y = y' =0.
By expression (4) of your attachment, when t' = 0, all the terms containing $sin(y\omega t')$ vanish and the term containing y' vanishes too, leaving

$$dx = dx'cos(0) = dx'$$.

Although that seems trivial, the same is not true for dy at time t'=0.
The dy equation in expression (4) becomes:

$$dy = \gamma dy' +R\gamma\omega dt'$$

 

[starthaus]

Sorry, I actually meant when $\theta = 0$, where $\omega=d\theta/dt$

It is perhaps time that you clear up your attachment - the first line of your equations [5] of "Uniformly Rotating Frames" does use the contentious 'mixing of frames' (dx/dt').

No, it doesn't. It contains the standard "chain rule" for calculating derivatives.

$$\frac {dx}{dt}=\frac{dx}{dt'} \frac{dt'}{dt}$$

This is standard math, not the type of frame-mixing that kev does.

It would also be immensely helpful to all around here (especially me as an engineer, no mathematician) if you would correct the typo and complete equations [6]...

I am quite sure that you and kev can complete expressions (6). I left them unfiinished on purpose, to lead you to the correct results and to teach you the correct way of solving this problem. If I give you the result ready made, kev will continue to find imaginary errors. Much better if I guide you to finding it.

 

[starthaus]

Jorrie obviously meant that when the angle (theta or phi or whatever you choose to call it) between the x axes of the two frames is zero and by the definition given in this link that you gave http://arxiv.org/PS_cache/gr-qc/pdf/9904/9904078v2.pdf this occurs when t' =0.


It does happen. At time t' = 0, the axes are aligned and x = x' = 0 and y = y' =0.
By expression (4) of your attachment, when t' = 0, all the terms containing $sin(y\omega t')$ vanish and the term containing y' vanishes too, leaving

$$dx = dx'cos(0) = dx'$$.

True, it happens in ONE PARTICULAR case. It does NOT happen in ANY other INFINITE number of cases. What does this tell you?

Although that seems trivial, the same is not true for dy at time t'=0.
The dy equation in expression (4) becomes:

$$dy = \gamma dy' +R\gamma\omega dt'$$

Yep, it NEVER happens. What does this tell you? Do you still claim that you can use the expressions derived from Lorentz transforms for translational motion?

 

[Dale]

I hesitate to get involved in a good sparring match, but I have an old Mathematica notebook on relativistic uniform circular motion using four-vectors. I could post the results here if that would be helpful in any way.

 

[starthaus]

I hesitate to get involved in a good sparring match, but I have an old Mathematica notebook on relativistic uniform circular motion using four-vectors. I could post the results here if that would be helpful in any way.

I think that it is probably best to let kev work out the answers on his own. Only this way will he learn, giving away the answers is not the best way to teach somebody. He has been given all the tools, it is now up to him to complete the derivation.

 

[yuiop]

I hesitate to get involved in a good sparring match, but I have an old Mathematica notebook on relativistic uniform circular motion using four-vectors. I could post the results here if that would be helpful in any way.

I would be very interested in your contribution. I think this thread is well overdue some input from the more knowledgeable members of this forum.

 

[yuiop]

I think that it is probably best to let kev work out the answers on his own. Only this way will he learn, giving away the answers is not the best way to teach somebody. He has been given all the tools, it is now up to him to complete the derivation.

This thread is not just about "teach kev a lesson". It's primary objective is to discuss the issues involved in relativistic centripetal force and acceleration in a way that might be useful for members of this forum. If that means I, you, Jorrie or anyone else learns something along the way, then all well and good. It is not just about your ego. A think a worthy secondary objective would be debug the document you presented in your blog, so that it might become a useful reference document for members of this forum. At this moment in time it far from status.

 

[yuiop]

$$dx = dx'cos(0) = dx'$$.

True, it happens in ONE PARTICULAR case. It does NOT happen in ANY other INFINITE number of cases. What does this tell you?

That tells me that the x and y components are changing over time, but because of the circular symmetry and because the rotation is uniform and constant, the angular velocity, centripetal force, centripetal acceleration and gamma factor all remain constant over time and for any any angle theta and those are the physical quantities jorrie and I are interested in. Your equations seem unable to determine those quantities.

Do you still claim that you can use the expressions derived from Lorentz transforms for translational motion?

Yes I do when we talk about MCIF's and proper acceleration as measured by an accelerometer. Your equations are about acceleration in terms of spatial displacement where a glass paperweight on your table has zero acceleration because its location is not changing over time, while an accelerometer will show the paperweight is accelerating despite the fact it appears to be stationary in your accelerating reference frame.


The reference paper that you lifted the initial eqations from http://arxiv.org/PS_cache/gr-qc/pdf/9904/9904078v2.pdf explicitly disagrees with your last statement when they say at (19):

In particular, at t′ = 0 these transformations ... coincide with the ordinary Lorentz boost at t′ = 0 for the velocity in the y-direction.

and later on when they say in Section 6:

Therefore, for a small range of values of t′, the transformations (6)-(7) can be approximated by the ordinary Lorentz boosts (see (19)). From this fact we conclude that if a moving rigid body is short enough, then its relativistic contraction in the direction of the instantaneous velocity, as seen from S, is simply given by L(t) = L′/(t), i.e., it depends only on the instantaneous velocity, not on its acceleration and rotation. (“Short enough” means that L′ ≪ c2/a′ k, where a′k is the component of the proper acceleration parallel to the direction of the velocity [11]).
By a similar argument we may conclude that an arbitrarily accelerated and rotating observer sees equal lengths of other differently moving objects as an inertial observer whose instantaneous position and velocity are equal to that of the arbitrarily accelerated and rotating observer.

 

[starthaus]

This thread is not just about "teach kev a lesson". It's primary objective is to discuss the issues involved in relativistic centripetal force and acceleration in a way that might be useful for members of this forum. If that means I, you, Jorrie or anyone else learns something along the way, then all well and good. It is not just about your ego.

It's not about my ego. It is about you stopping trying to find errors where they don't exist and starting to do a couple of differentiations such that you find the answer for yourself. (the first differentiation is already done, you only need to complete the second order).

A think a worthy secondary objective would be debug the document you presented in your blog,

There is no "debug". There is rolling up your sleeves and calculating how acceleration transforms under rotation using the appropiate Lorentz transforms.
I will give you a hint: the method that you need to apply is identical to the one one uses to derive acceleration from the Lorentz transforms for translation. Except that you need to use the Lorentz transforms for rotation that have been provided for you. Rather than arguing for days about imaginary errors you could have obtained the answer by now.

 

[yuiop]

Your exercise is to express $$\frac {d^2x}{dt^2}$$ as a function of $$\frac {d^2x'}{dt'^2}$$. I gave you all the tools to do that correctly.

You have not obtained $$\frac {d^2x'}{dt'^2}$$ in your document. All you have done is taken the derivative of x with respect to time twice and then transformed the result by converting dt to dt' using dt = gamma(dt'), What you have failed to do, is convert dx to dx' using dx = f(dx') where f is the long function inplicit in expression (4) of your document.

I am quite sure that you and kev can complete expressions (6). I left them unfiinished on purpose, to lead you to the correct results and to teach you the correct way of solving this problem. If I give you the result ready made, kev will continue to find imaginary errors. Much better if I guide you to finding it.

It is easy to complete expression (6) because it is obvious from the terms you have completed and from the method you use in obtaining expression (5), that all you are doing is dividing each term by $\gamma dt'$. The completed expressions are (as far as I can tell):

$$\frac{d^2x}{dt^2}=\gamma^{-2}\left(\frac{d^2x'}{dt'^2}\cos(\gamma\omega t')-\gamma\omega\sin(\gamma\omega t') \frac{dx'}{dt'}-\gamma\frac{d^2y'}{dt'^2}\sin(\gamma\omega t')-\frac{dy'}{dt'}\gamma^{2}\omega\cos(\gamma\omega t') - \frac{d}{dt'}R\gamma\omega\sin(\gamma\omega t')\right)$$

$$\frac{d^2y}{dt^2}=\gamma^{-2}\left(\frac{d^2x'}{dt'^2}\sin(\gamma\omega t')+\gamma\omega\cos(\gamma\omega t') \frac{dx'}{dt'}+\gamma\frac{d^2y'}{dt'^2}\cos(\gamma\omega t')-\frac{dy'}{dt'}\gamma^{2}\omega\sin(\gamma\omega t') + \frac{d}{dt'}R\gamma\omega\cos(\gamma\omega t')\right)$$

If that is not what you intended, I have done the hard work of formatting the tex and all you have to do is edit it.

Note that for constant uniform rotation, the last term of each expression containg R probably vanishes, because none of the quantities $R,\gamma,\omega$ are changing over time.

Also note that neither jorrie or myself claim to be expert mathematicians and it would be helpful to both of us if you would break down how you get from (1) to (4). After all, you claim your goal is educate me.

 

[starthaus]

You have not obtained $$\frac {d^2x'}{dt'^2}$$ in your document. All you have done is taken the derivative of x with respect to time twice and then transformed the result by converting dt to dt' using dt = gamma(dt'), What you have failed to do, is convert dx to dx' using dx = f(dx') where f is the long function inplicit in expression (4) of your document.

You need to stop trying to find fault with what I am trying to teach you. The method is showing how to find $$\frac {d^2x}{dt^2}$$ as a function of $$\frac {d^2x'}{dt'^2}$$, $$\frac {dx'}{dt'}$$, etc. Looks that after a lot of prodding, you went ahead and you did it yourself.

It is easy to complete expression (6) because it is obvious from the terms you have completed and from the method you use in obtaining expression (5), that all you are doing is dividing each term by $\gamma dt'$. The completed expressions are (as far as I can tell):

$$\frac{d^2x}{dt^2}=\gamma^{-2}\left(\frac{d^2x'}{dt'^2}\cos(\gamma\omega t')-\gamma\omega\sin(\gamma\omega t') \frac{dx'}{dt'}-\gamma\frac{d^2y'}{dt'^2}\sin(\gamma\omega t')-\frac{dy'}{dt'}\gamma^{2}\omega\cos(\gamma\omega t') - \frac{d}{dt'}R\gamma\omega\sin(\gamma\omega t')\right)$$

$$\frac{d^2y}{dt^2}=\gamma^{-2}\left(\frac{d^2x'}{dt'^2}\sin(\gamma\omega t')+\gamma\omega\cos(\gamma\omega t') \frac{dx'}{dt'}+\gamma\frac{d^2y'}{dt'^2}\cos(\gamma\omega t')-\frac{dy'}{dt'}\gamma^{2}\omega\sin(\gamma\omega t') + \frac{d}{dt'}R\gamma\omega\cos(\gamma\omega t')\right)$$

Good, this is correct.

Note that for constant uniform rotation, the last term of each expression containg R probably vanishes,

False, it doesn't "vanish". This is basic calculus, remember?

 

[Dale]

I would be very interested in your contribution.

Since the vote is split 50/50 I will appoint myself vice president and cast the deciding vote in favor of posting it

So throughout this I will use the convention that the time coordinate is the first coordinate and that timelike intervals squared are positive. So given a wheel centered at the origin and rotating about the z axis in a counter clockwise direction we can describe the worldline for any particle on the rim by:
$$\mathbf s =(ct,\; r \; cos(\phi + \omega t),\; r \; sin(\phi + \omega t),\; 0)$$

The four-velocity is given by:
$$\mathbf u=c \frac{d\mathbf s/dt}{|d\mathbf s/dt|}=(\gamma c,\; -\gamma r \omega \; sin(\phi + \omega t),\; \gamma r \omega \; cos(\phi + \omega t),\; 0)$$
where |x| indicates the Minkowski norm of the four-vector x and $$\gamma=(1-\frac{r^2 \omega^2}{c^2})^{-1/2}$$

The four-acceleration is given by:
$$\mathbf a=c \frac{d\mathbf u/dt}{|d\mathbf s/dt|} = (0,\; -\gamma^2 r \omega^2 \; cos(\phi + \omega t),\; -\gamma^2 r \omega^2 \; sin(\phi + \omega t),\; 0 )$$

So the magnitude of the four-acceleration (which is frame invariant and equal to the magnitude of the proper acceleration) is given by:
$$|\mathbf a|^2=-\gamma^4 r^2 \omega^4$$

That is what I had in my Mathematica file, but I certainly can go from there or explain things if either of you have questions or requests.

 

[starthaus]

Since the vote is split 50/50 I will appoint myself vice president and cast the deciding vote in favor of posting it

So throughout this I will use the convention that the time coordinate is the first coordinate and that timelike intervals squared are positive. So given a wheel centered at the origin and rotating about the z axis in a counter clockwise direction we can describe the worldline for any particle on the rim by:
$$\mathbf s =(ct,\; r \; cos(\phi + \omega t),\; r \; sin(\phi + \omega t),\; 0)$$

The four-velocity is given by:
$$\mathbf u=c \frac{d\mathbf s/dt}{|d\mathbf s/dt|}=(\gamma c,\; -\gamma r \omega \; sin(\phi + \omega t),\; \gamma r \omega \; cos(\phi + \omega t),\; 0)$$
where |\mathbf x| indicates the Minkowski norm of the four-vector x and $\gamma=(1-\frac{r^2 \omega^2}{c^2})^{-1/2}$

The four-acceleration is given by:
$$\mathbf a=c \frac{d\mathbf u/dt}{|d\mathbf s/dt|} = (0,\; -\gamma^2 r \omega^2 \; cos(\phi + \omega t),\; -\gamma^2 r \omega^2 \; sin(\phi + \omega t),\; 0 )$$

So the magnitude of the four-acceleration (which is frame invariant and equal to the magnitude of the proper acceleration) is given by:
$$|\mathbf a|^2=-\gamma^4 r^2 \omega^4$$

Careful, see here. There is the additional condition $$\gamma=1$$ that must not be overlooked.

That is what I had in my Mathematica file, but I certainly can go from there or explain things if either of you have questions or requests.

We've been over this, this is exactly the same as pervect's post. The subject of the discussion is the appropiate transforms between frames (see post #3). Lorentz transforms for translation are not appropiate for describing rotation. kev is almost done with learning how to use the Lorentz transforms for rotation.

 

[Dale]

Careful, see here. There is the additional condition $$\gamma=1$$ that must not be overlooked.

No. Since,$$\gamma=(1-\frac{r^2 \omega^2}{c^2})^{-1/2}$$ the condition $\gamma=1$ would imply $\omega=0$ which is not true in general.

We've been over this, this is exactly the same as pervect's post. The subject of the discussion is the appropiate transforms between frames (see post #3). Lorentz transforms for translation are not appropiate for describing rotation. kev is almost done with learning how to use the Lorentz transforms for rotation.

I deliberately didn't do any Lorentz transforms nor did I look at any other reference frame.

Although a boost is not a spatial rotation you can always find a momentarily co-moving inertial frame (MCIF) for any event on any particle's worldline. This frame will be related to the original frame via a Lorentz transform. In such a frame the particle will only be at rest for a moment, hence the term "momentarily co-moving". So you cannot use a Lorentz transform to find a (non-instantaneous) rest frame for a particle in general, but there is a lot of useful parameters you can obtain from the MCIF. I don't know how kev was using it.

 

[Jorrie]

Since the vote is split 50/50 I will appoint myself vice president and cast the deciding vote in favor of posting it

Thanks DaleSpam, I should have voted for it - I have been asking Starthaus for his treatment before.

Anyway, it seems that your magnitude of the proper acceleration answers the original question of the OP: the relativistic centrifugal force, measured in the way that kev described.

To me it does not seem as if Starthaus's treatment answered that question, but rather a different one on coordinate transformations for rotating frames. I am sure that it is useful in its own right and can possibly be extended to answer the original question.

 

[starthaus]

I deliberately didn't do any Lorentz transforms nor did I look at any other reference frame.

True, neither did you answer the question in discussion. You found the expression for four-acceleration (something that we already knew) but you did not answer the question as to how centripetal force transforms. In order to do that , you need to start from the Lorentz transforms for rotating frames.

Although a boost is not a spatial rotation you can always find a momentarily co-moving inertial frame (MCIF) for any event on any particle's worldline. This frame will be related to the original frame via a Lorentz transform. In such a frame the particle will only be at rest for a moment, hence the term "momentarily co-moving". So you cannot use a Lorentz transform to find a (non-instantaneous) rest frame for a particle in general, but there is a lot of useful parameters you can obtain from the MCIF.

Sure but why use this hack when you are being shown how to use the appropiate Lorentz transforms?

I don't know how kev was using it.

With lots of mistakes.

 

[starthaus]

Thanks DaleSpam, I should have voted for it - I have been asking Starthaus for his treatment before.

Physics is not decided by voting.

Anyway, it seems that your magnitude of the proper acceleration answers the original question of the OP: the relativistic centrifugal force, measured in the way that kev described.

No, it doesn't. Look at post #3, the issue being contested is usage of the appropiate transforms in finding out the way the centripetal force transforms between frames.

To me it does not seem as if Starthaus's treatment answered that question, but rather a different one on coordinate transformations for rotating frames. I am sure that it is useful in its own right and can possibly be extended to answer the original question.

kev is not finished with his exercise , he is just one step away from finishing it. If he corrects his last mistake, he's one step away from getting the correct answer using the correct formalism.
Would you like to finish it for him? It is a matter of doing the derivative correctly and setting some conditions.

 

[starthaus]

To me it does not seem as if Starthaus's treatment answered that question, but rather a different one on coordinate transformations for rotating frames. I am sure that it is useful in its own right and can possibly be extended to answer the original question.

The method I am showing solves a much more general question. I will show you how it also solves the question asked by the OP , as a trivial particular case. Just stick with the approach and you will find out the correct answers obtained through the appropiate formalisms.

 

[starthaus]

No. Since,$$\gamma=(1-\frac{r^2 \omega^2}{c^2})^{-1/2}$$ the condition $\gamma=1$ would imply $\omega=0$ which is not true in general.

No, it doesn't mean that, it only means that the comoving frame (for which you are calculating the proper acceleration) is characterized by $\omega=0$ . Because the frame is co-moving with the rotating object.
So, you should have obtained the proper acceleration magnitude as:

$$a=r\omega^2$$

 

[yuiop]

OK, if I understand correctly, because of the t' contained in the cos and sin functions in the last terms, the expressions for (6) should be:

$$\frac{d^2x}{dt^2}=\gamma^{-2}\left(\frac{d^2x'}{dt'^2}\cos(\gamma\omega t')-\gamma\omega\sin(\gamma\omega t') \frac{dx'}{dt'}-\gamma\frac{d^2y'}{dt'^2}\sin(\gamma\omega t')-\frac{dy'}{dt'}\gamma^{2}\omega\cos(\gamma\omega t') - R\gamma^2\omega^2\cos(\gamma\omega t')\right)$$ (Eq6x)

$$\frac{d^2y}{dt^2}=\gamma^{-2}\left(\frac{d^2x'}{dt'^2}\sin(\gamma\omega t')+\gamma\omega\cos(\gamma\omega t') \frac{dx'}{dt'}+\gamma\frac{d^2y'}{dt'^2}\cos(\gamma\omega t')-\frac{dy'}{dt'}\gamma^{2}\omega\sin(\gamma\omega t') - R\gamma^2\omega^2\sin(\gamma\omega t')\right)$$ (Eq6y)

Is that correct? As I said before, I am no math expert. Should the same be done for the other terms in the expression?

If they are correct, then when t' = 0, $\sin(\gamma\omega t') = 0$ and $\cos(\gamma\omega t') = 1$ and the equations simplify to:

$$a_x = \frac{d^2x}{dt^2}=\gamma^{-2}\left(\frac{d^2x'}{dt'^2}-\frac{dy'}{dt'}\gamma^{2}\omega- R\gamma^2\omega^2\right)$$ (Eq7x)

$$a_y = \frac{d^2y}{dt^2}=\gamma^{-2}\left(\gamma\omega \frac{dx'}{dt'}+\gamma\frac{d^2y'}{dt'^2}\right)$$ (Eq7y)

Now if the particle is stationary in the rotating frame and located on the y'=0 axis, which is the radial line from the centre of rotation to the particle, the result is:

$$a_x = \frac{d^2x}{dt^2}= -R\omega^2\right)$$

$$a_y = \frac{d^2y}{dt^2}= 0$$

which is a good result, because that is what we expect in the non-rotating lab frame. This is the coordinate acceleration.

Equations (7x) and (7y) can be rearranged to give:

$$a'_x = \frac{d^2x'}{dt'^2}=\gamma^{2}\left(\frac{d^2x}{dt^2}+\frac{dy'}{dt'}\omega+ R\omega^2\right)$$

$$a'_y = \frac{d^2y'}{dt'^2}=\gamma\frac{d^2y}{dt^2}-\omega \frac{dx'}{dt'}$$

If the particle is released so that it flies off in a straight line from the point of view of the lab frame at time t’=0, its initial velocity in the rotating frame was zero so that upon release $dx\prime/dt\prime \approx 0$ and $dy\prime/dt\prime \approx 0$ then:

$$a'_x = \frac{d^2x'}{dt'^2} =\gamma^{2}\left(\frac{d^2x}{dt^2}+\frac{dy'}{dt'}\omega+ R\omega^2\right) =\gamma^{2}\left(0+\frac{dy'}{dt'}\omega+ R\omega^2\right) \approx \gamma^{2}\left(R\omega^2\right)$$

$$a'_y = \frac{d^2y'}{dt'^2}=\gamma\frac{d^2y}{dt^2}-\omega \frac{dx'}{dt'} =0-\omega \frac{dx'}{dt'} \approx 0$$

So in the lab frame the centripetal acceleration of the particle when it is traveling in a circle is $R\omega^2$ and when it is released the acceleration of the particle according to the observer in the rotating frame is initially $\gamma^{2} R\omega^2$. This magnitude of this measurement coincides with with the magnitude of the four acceleration given by Pervect and Dalespam (Thanks for the imput by the way ), which is the proper acceleration as measured by an accelerometer.

Note:The x and x' axes always pass through the center of rotation so $a_x$ and $a_x'$ represent the inward pointing centripetal acceleration at time t'=0, which these calculations are based on. $a_y$ is the angular acceleration directed tangentially.

Conclusion

Coordinate centripetal acceleration:

$$a_x = R\omega^2$$ (Proven by Starthaus)

Proper centripetal acceleration:

As measured by an accelerometer in the rotating frame:

$$a\prime_x = \gamma^2 R\omega^2$$ (Proven by Dalespam and Pervect)

As measured by initial “fall rate acceleration” in the rotating frame:

$$a\prime_x \approx \gamma^2 R\omega^2$$ (Proven by Starthaus)

The above is basically in agreement with everything I said in #1 but the methods of measurement have been more carefully defined now.

This has significance because it tells us how a particle behaves in a gravitational field, by applying the equivalence principle.

 

[Dale]

you need to start from the Lorentz transforms for rotating frames.

I have never heard of the Lorentz transform for rotating frame. AFAIK the Lorentz transform transforms between different inertial frames, not between an inertial frame and a rotating frame.

 

[Dale]

No, it doesn't mean that, it only means that the comoving frame (for which you are calculating the proper acceleration) is characterized by $\omega=0$ . Because the frame is co-moving with the rotating object.
So, you should have obtained the proper acceleration magnitude as:

$$a=r\omega^2$$

The Minkowski norm of the four-acceleration is frame invariant. So it does not matter which frame you calculate it in; it will be the same in all frames. If those factors of $1-r^2\omega^2/c^2$ appear in one frame they must appear in all frames.

 

[starthaus]

OK, if I understand correctly, because of the t' contained in the cos and sin functions in the last terms, the expressions for (6) should be:

$$\frac{d^2x}{dt^2}=\gamma^{-2}\left(\frac{d^2x'}{dt'^2}\cos(\gamma\omega t')-\gamma\omega\sin(\gamma\omega t') \frac{dx'}{dt'}-\gamma\frac{d^2y'}{dt'^2}\sin(\gamma\omega t')-\frac{dy'}{dt'}\gamma^{2}\omega\cos(\gamma\omega t') - R\gamma^2\omega^2\cos(\gamma\omega t')\right)$$ (Eq6x)

$$\frac{d^2y}{dt^2}=\gamma^{-2}\left(\frac{d^2x'}{dt'^2}\sin(\gamma\omega t')+\gamma\omega\cos(\gamma\omega t') \frac{dx'}{dt'}+\gamma\frac{d^2y'}{dt'^2}\cos(\gamma\omega t')-\frac{dy'}{dt'}\gamma^{2}\omega\sin(\gamma\omega t') - R\gamma^2\omega^2\sin(\gamma\omega t')\right)$$ (Eq6y)

Is that correct?

Correct. Finally.

If they are correct, then when t' = 0, $\sin(\gamma\omega t') = 0$ and $\cos(\gamma\omega t') = 1$ and the equations simplify to:

Incorrect. In the comoving frame you need to realize that $$\frac{dx'}{dt'}=0$$ and $$\frac{dy'}{dt'}=0$$. Try continuing the calculations from this hint.

 

[starthaus]

I have never heard of the Lorentz transform for rotating frame. AFAIK the Lorentz transform transforms between different inertial frames, not between an inertial frame and a rotating frame.

I gave a couple of references earlier in the thread. Here they are again:

1. R. A. Nelson, "Generalized Lorentz transformation for an accelerated, rotating frame of reference", J. Math. Phys. 28, 2379 (1987);

2. H.Nikolic, “Relativistic contraction and related effects in non-inertial frames”, Phys.Rev. A, 61, (2000)

 

[starthaus]

The Minkowski norm of the four-acceleration is frame invariant. So it does not matter which frame you calculate it in; it will be the same in all frames. If those factors of $1-r^2\omega^2/c^2$ appear in one frame they must appear in all frames.

No argument. The argument is about the fact that proper acceleration is DIFFERENT from four-acceleration. It differs exactly by the $$\gamma$$ factor. What you calculated iis the four-acceleration, not the proper acceleration. But, we are digressing. The discussion is about the correct usage of Lorentz transforms. Please be patient, kev is one step away from finishing the computations.